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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 12$ and $Q = \cos \left(\frac{\p

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There exists a real number $\beta$ such that the line $y = \beta$ intersects the curve $y = y(x)$ at infinitely many points

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 12$ and $Q = \cos \left(\frac{\pi}{12} xight)$.The integrating factor $(I.F.)$ is $e^{\int 12 dx} = e^{12x}$.The general solution is $y \cdot e^{12x} = \int e^{12x} \cos \left(\frac{\pi}{12} xight) dx + C$.Using the formula $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx))$,we get:$y \cdot e^{12x} = \frac{e^{12x}}{12^2 + (\frac{\pi}{12})^2} \left(12 \cos \left(\frac{\pi}{12} xight) + \frac{\pi}{12} \sin \left(\frac{\pi}{12} xight)ight) + C$.Simplifying,$y = \frac{1}{144 + \frac{\pi^2}{144}} \left(12 \cos \left(\frac{\pi}{12} xight) + \frac{\pi}{12} \sin \left(\frac{\pi}{12} xight)ight) + C e^{-12x}$.Given $y(0) = 0$,we have $0 = \frac{12}{144 + \frac{\pi^2}{144}} + C$,so $C = -\frac{12}{144 + \frac{\pi^2}{144}}$.Thus,$y(x) = \frac{1}{144 + \frac{\pi^2}{144}} \left(12 \cos \left(\frac{\pi}{12} xight) + \frac{\pi}{12} \sin \left(\frac{\pi}{12} xight) - 12 e^{-12x}ight)$.As $x 	o \infty$,$e^{-12x} 	o 0$,so $y(x)$ approaches a periodic function $f(x) = A \cos \left(\frac{\pi}{12} x - \phiight)$.Since $y(x)$ approaches a periodic function,for a value $\beta$ within the range of this periodic function,the line $y = \beta$ will intersect the curve $y = y(x)$ at infinitely many points as $x 	o \infty$. Thus,statement $C$ is $TRUE$.

For $x \in R$,let the function $y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 12y = \cos \left(\frac{\pi}{12} x\right)$ with $y(0) = 0$. Then,which of the following statements is/are $TRUE$?

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